Timeline for Cohomology of Groups at Gregory Berhuy's Book
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 25, 2012 at 13:29 | answer | added | Ronnie Brown | timeline score: -4 | |
| Jun 25, 2012 at 10:00 | comment | added | Zachi Evenor | Thanks. Now that this understood, I got the rest (I recall the since the sequence is exact $g \circ f = 1$ the trivial map. | |
| Jun 24, 2012 at 20:23 | comment | added | Mikhail Borovoi | In this book on page 52, $\beta_\sigma$ is SOME preimage of $\gamma_\sigma$. Although $\gamma$ is a cocycle, $\beta$ in general is NOT a cocycle, and this is the point! It is a cochain, whose coboundary lives in $f(A)$ (because $\gamma$ is a cocycle), which gives us a 2-cocycle in $A$. The cohomology class of this cocycle is, by definition, $\delta^1(\gamma)$. | |
| Jun 24, 2012 at 17:15 | comment | added | Zachi Evenor | @stankewicz See my last comment, it may help. | |
| Jun 24, 2012 at 12:05 | comment | added | Zachi Evenor | I'm not sure. He first states that $$g( \beta_\sigma (\sigma \cdot \beta_\tau ) \beta_{\sigma \tau}^{-1}) = \gamma_\sigma \sigma \cdot \gamma_\tau \gamma_{\sigma \tau}^{-1} = 1 $$ and then deduces from that, that $$ \beta_\sigma (\sigma \cdot \beta_\tau ) \beta_{\sigma \tau}^{-1} = f( \alpha_{\sigma , \tau})$$ (I'm quoting: " = 1, so $\beta ... = f( \alpha_{\sigma , \tau})$ for some unique $\alpha_{\sigma , \tau} \in A$). However, what about the general case? Why it is not =1 regardless the $f,g$ homomorphisms? | |
| Jun 24, 2012 at 11:36 | history | edited | stankewicz |
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| Jun 24, 2012 at 11:36 | comment | added | stankewicz | It seems to me like he's using the fact that $g(\beta_\sigma(\sigma \cdot \beta_\tau)\beta_{\sigma\tau}^{-1}) =1$ to say that \beta_\sigma(\sigma \cdot \beta_\tau)\beta_{\sigma\tau}^{-1} is in the image of $f$, which is of course true by the assumption that we have a short exact sequence of $G$-groups. | |
| Jun 24, 2012 at 11:27 | history | asked | Zachi Evenor | CC BY-SA 3.0 |